What is the integral of 4/(4x+1) ?

The integral of 4/(4x+1) is ln|4x+1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4/(4x+1)

\int \frac{4}{4 x + 1} d x

ln ∣ 4 x + 1 ∣ + C

Solution steps

\int \frac{4}{4 x + 1} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{1}{4 x + 1} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

= 4 \cdot \int \frac{1}{4 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{4} ln ∣ u ∣

Substitute back

u = 4 x + 1

= 4 \cdot \frac{1}{4} ln ∣ 4 x + 1 ∣

Simplify

4 \cdot \frac{1}{4} ln ∣ 4 x + 1 ∣ : ln ∣ 4 x + 1 ∣

= ln ∣ 4 x + 1 ∣

Add a constant to the solution

= ln ∣ 4 x + 1 ∣ + C

e^{x} \frac{d y}{d x} = 2 x

(t - 1) cos (4 (t - 1)) + t^{2}

f (x) = \frac{x ^{2}}{4 - cos ( x )}

\int sin^{2} (\frac{x}{2}) cos^{2} (\frac{x}{2}) d x

f (x) = \frac{1 - x}{1 + x}